# Thread: Finding the roots through Newtons method

1. ## Finding the roots through Newtons method

In each of the following items approximate the zeros of $\displaystyle f$ using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001

The Newton iteration: $\displaystyle x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}$

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till $\displaystyle |x_{n+1}-x_n|\leq{0.001}|$, or if I should apply some of this:

$\displaystyle k_1>0$, $\displaystyle |f'(x)|\geq{k_1}$ and $\displaystyle |f''(x)|\leq{}k_2$ for all $\displaystyle x\in{[b]}$, then:

$\displaystyle |x_{n+1}-r|<\displaystyle\frac{k_2}{2k_1}|x_n-r|^2$

If $\displaystyle r\in{}[r-\delta,r+\delta]\subset{[a,b]}$, and $\displaystyle \delta<2(\displaystyle\frac{k_1}{k_2})$

$\displaystyle |x_{n+1}-r|<\displaystyle\frac{2k_1}{k_2}(\displaystyle\fra c{\delta}{\displaystyle\frac{2k_1}{k_2}})^2n$

2. Originally Posted by Ulysses
In each of the following items approximate the zeros of $\displaystyle f$ using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001

The Newton iteration: $\displaystyle x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}$

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till $\displaystyle |x_{n+1}-x_n|\leq{0.001}$,
Well, yes, I think this is very clearly what you are asked to do.

or if I should apply some of this: <more complicated stuff>
I know that in order to show that Newton's method converges "quadratically" (that is, that the number of valid digits doubles with each iteration), if it converges at all, one does have to do some more fancy guesswork like this. But I think it is obvious that the text of your exercise does not require you to handle such technicalities.